Question 616970: Can you please give me the answer to this equation, and explain how the answer was derived?
If 5x + x² > 100, then x is not
Thank you.
Found 2 solutions by ewatrrr, solver91311:
Answer by ewatrrr(24785)   (Show Source):
You can put this solution on YOUR website!
Hi,
If 5x + x² > 100 then x^2 + 5x -100 > 0 and x is not in the interval [-12.81,7.81]
f(x) = x^2 + 5x -100 
x-intercepts: (7.81,0) and (-12.81)
Answer by solver91311(24713)  (Show Source):
You can put this solution on YOUR website!
It is impossible to do what you ask. You can't solve an equation that doesn't exist. What you have provided is an inequality, NOT an equation. On the other hand, you need to turn it into an equation in order to determine the solution set interval.
Step 1: Replace the inequality sign with an equals sign.
Step 2: Put the equation into standard quadratic form.
Step 3: Evaluate the discriminant, i.e.,  , to determine that the quadratic does not factor over the rationals.
Step 4: Use the quadratic equation to find the conjugate pair of irrational roots of the equation.
Step 5: Divide the  -axis into three regions,  up to the smaller value root, the smaller value root to the larger value root, and the larger value root to  .
Step 6: Select any one of the three intervals, and select a value from that interval that is NOT an endpoint value. Substitute this value into the original inequality and do the arithmetic.
Step 7: If the result of step 6 is a true statement, then the interval selected is contained in the solution set of the inequality. Otherwise not.
Step 8: Repeat steps 6 and 7 for each of the other intervals.
Use the results to fully describe the solution set. Generally, interval notation is best.
John
My calculator said it, I believe it, that settles it
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